0
\$\begingroup\$

Attached is the snapshot of the circuit I am working with.

enter image description here

My aim is to find the minimum input voltage to turn on the transistor (T)

With the following assumption, voltage drop across the resistor R1 is very small,

Vin > 2.7V + 0.6V ( Zener voltage + drop across V_BE)

That is, minimum input voltage should be minimum of 3.3V to turn on the transistor (T).

Questions

  1. Can someone verify the calculation?

  2. How will calculation changes if drop across R1 also need incorporate?

\$\endgroup\$
  • 1
    \$\begingroup\$ Have you defined what threshold "turn on" is? \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 21 '17 at 13:12
  • \$\begingroup\$ @IgnacioVazquez-Abrams: yes, 3.2V \$\endgroup\$ – vt673 Jan 21 '17 at 13:17
  • \$\begingroup\$ Your 2.7 V + 0.6 V is correct. At that point current starts to flow, so (ideal world) at 3.3 V the current is still zero, at 3.30001 V a small current starts to flow. Now think about what the voltage drop will be in both cases. \$\endgroup\$ – Bimpelrekkie Jan 21 '17 at 13:18
  • 3
    \$\begingroup\$ By the way, this is a horrible example, probably formulated by someone with very little real-world design experience. If you doubt that, assemble the parts and test it. \$\endgroup\$ – Spehro Pefhany Jan 21 '17 at 13:30
  • 2
    \$\begingroup\$ @vt673 Vz is specified at a relatively high current (typically 5-20mA for a 2.7V zener) compared to the base current required to saturate a transistor as shown (maybe a few uA). 2.7V zeners will conduct a few uA at much lower voltage than 2.7V- likely well under 1.0V. A base-emitter resistor should be used and a higher voltage zener (such as 12V) would be better since they use a different breakdown mechanism than << 5V zeners and are 'sharper'. \$\endgroup\$ – Spehro Pefhany Jan 21 '17 at 15:04
3
\$\begingroup\$

Can someone verify the calculation?

the calculation is generally correct, but dependent on your definition of "turn on". for most analog applications, people use "turn on" to mean "starting to turn on". and for digital applications, people tend to use "fully turn on" instead.

as Vbe is not a constant, nor the reverse voltage of the zener, the figure you obtained is just a rough estimate, and should be adjusted further for specific applications if called for.

How will calculation changes if drop across R1 also need incorporate?

you should calculate the Ic first, and then apply a Ib based on the assumed beta of the transistor. That current will cause a voltage drop over R1.

If you use typical values, you will find that that voltage drop is quite small vs. Vbe + Vzfwd, thus most times ignored.

However, there can be cases where the voltage drop over R1 is significant. A prudent design practice is to first assume that it is insignificant, go through your calculations, obtain Ib and then confirm that it is indeed significant.

edit1:

I'm amending this in response to the answer provided by Jack Creasey below. Jack had concluded that the Ib would be 16ua max, based on an estimated beta of 50.

I pointed out that beta is really a concept for linear applications (where the associated Vce is much higher than in a switching application here).

Here is a simulation I put together quickly - it mimics the circuit provided by the OP, with the exception that I used a 4.7v zener, not a 2.7v zener. the use of a 2.7v zener would have increased the Ib by 200ua in this case, with minimum changes in Ic.

enter image description here

the blue line in the plot is Ib and it reaches 500ua at 10v input, with Ic at 800ua -> a beta of 1.6.

hope it helps.

if anyone is interested, I can simulate the case with a 2.7v zener, :)

edit2: due to popular demand, here is the same circuit, using a 2.7v zener.

Ib tops out at about 680ua, vs. 500ua with a 4.7v zener, Ic unchanged, just as we expected.

enter image description here

again, the point I have been trying to emphasize is that beta is a useful concept for linear applications and has no meaning in a switching application, as shown here.

\$\endgroup\$
1
\$\begingroup\$

Your calculations are incorrect for anything less than ideal components. The Zener characteristics look like this:

enter image description here

The BJT characteristics look like this:

enter image description here

Let's deal with some practical components:
If the Zener were a BZX79-B/C2V7 the leakage current could be as high as 20 uA @1 V.
If the transistor were a 2N2222 then we might expect the minimum Hfe would be around 50.

It's hard to know what the collector voltage is because you show a DC coupled restive divider of 4.3k and Rl (no value for RL). So I'll ignore RL for the moment.

For the Transistor to be on the collector current would be around (ignoring VCE(sat)) 0.8 mA and with an Hfe of 50 that implies a base current of around 16 uA maximum.
Have a look at the datasheet (Table 3) to see the B/C transfer characteristics.

enter image description here

Notice that the base current required to turn on the transistor could be less than the leakage current of the Zener!!
Notice in the graph that they suggest that the transistor may be on between 6-10 uA for 1 mA collector current suggesting an Hfe of over 150 for the device they used, and this will vary over a range for the transistors you might purchase.

So let's work out the input voltage based on a leakage current of the Zener of 20 uA at 1 V, and V(BE) for the transistor is 0.6 V.
20 UA through the 10k resistor results in a voltage drop of 0.2 V, so the input voltage would be 0.2 + 1 + 0.6 = 1.8 V !!!
Far from what you might expect.

If you want to be able to design at even a very simplified level, you need to understand the characteristics of your components.
In this case your very low currents don't help. You need to be able to ensure you design current flows in a DC circuit such as this to be immune to things such as component leakage currents.

Here you might put a resistor between the base and emitter of the transistor to shunt leakage current as shown below (though I used a 3.3V zener in this case). You can run the DC simulation to show the turn on/off of the transistor with input signal.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ > For the Transistor to be on the collector current would be around (ignoring VCE(sat)) 0.8 mA and with an Hfe of 50 that implies a base current of around 16 uA maximum. this is a switching application so you can easily over-drive the base. the base current here would be much higher than 16ua - you can either simulate or build a real circuit to verify that. \$\endgroup\$ – dannyf Jan 21 '17 at 21:17
  • \$\begingroup\$ @dannyf, whether the transistor is designed for switching is truly irrelevant. Look at the simulation and you can see that the switch slowly turns on as Ib increases, just as you'd expect. With an analog drive such as this you expect analog results. You could change the input to a square wave to overdrive the base but you still need to be out of the leakage current range. At minimum you need all the zener leakage current diverted. So if Vbe is 0.6 and leakage is 20 uA you need a maximum of 30k Ohm across the transistor BE. Typically it would be at least 10* --> 3k Ohms \$\endgroup\$ – Jack Creasey Jan 21 '17 at 22:04
  • \$\begingroup\$ @dannyf. The question was however, "What is the minimum voltage to turn on the transistor" ....so no overdrive, simply what voltage is required from a simple voltage generator. I think I answered that, though there is clearly no definitive number that could be calculated without having the detailed characteristics of both the zener and transistor available. \$\endgroup\$ – Jack Creasey Jan 21 '17 at 22:07
0
\$\begingroup\$

The transistor will function as an amplifer, running at nanoAmps or at microamps or at milliAmps, or at AMPS for audio power & L-band long-pulse-radar.

Between nanoAmps (I ignored the picoamps and femtoamps regions, because of external leakages) and Amps, we have 10^9 range or 9 decades of current.

With 58 milliVolts change in Vbe for each decade, we only need 522 milliVolts change in Vbe.

So where is the transistor on? I eventually learned to use "transconductance" to design amplifiers, and I use emitter resistors to set the currents. After lotta painful design failures as a student.

\$\endgroup\$
  • \$\begingroup\$ Yes - it is really a matter of DEFINITION. Lets take a simple example: The pn diode (because the B-E junction behaves like a diode). Some people consider the diode to be "on" if the forward bias is 0.7 volts. But this is (a) a raugh estimate and (b) nothing else than a definition. There will be a current even for a voltage of 0.1 volt. And the same applies, in principle, to ther BJT. We normally assume 0.65..0.7 volts for collector currents in the lower mA range - however, as outlined above, there will be a collector current also in the µA range. So - is there a "physical" theshold ? \$\endgroup\$ – LvW Feb 2 '17 at 7:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.